有向图的欧拉路径POJ2337

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
using namespace std;
const int maxn = 1100;
struct Edge {
    int to, next;
    int id;
}edge[maxn * 2];
int tot, head[maxn];
int in[maxn], out[maxn];
int F[maxn];
int st;
bool vis[30];
string str[maxn];
void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
    memset(F, -1, sizeof(F));
}
void addedge(int u, int v, int id)
{
    edge[tot].to = v;
    edge[tot].next = head[u];
    edge[tot].id = id;
    head[u] = tot++;
}
int Find(int x)
{
    if (F[x] == -1) return x;
    return F[x] = Find(F[x]);
}
void Union(int x, int y)
{
    int tx = Find(x);
    int ty = Find(y);
    if (tx != ty)
        F[tx] = ty;
}
bool check(int s)
{
    int in1 = 0, out1 = 0;
    for (int i = 1; i <= 26; i++)//判断出入度关系
    {
        if (in[i] == out[i]) continue;
        else if (in[i] - out[i] == 1) in1++;
        else if (out[i] - in[i] == 1) out1++, st = i;//如果有出度比入度大1的，说明是欧拉通路，起点只能是那个出度比入度大1的那个点
        else return false;
    }
    //printf("in1 = %d, out1 = %d\n", in1, out1);
    if (!(in1 == 1 && out1 == 1) && !(in1 == 0 && out1 == 0)) return false;
    for (int i = 1; i <= 26; i++)//判断弱联通
        if (vis[i] && Find(i) != Find(s))
            return false;
    return true;
}
bool vis2[maxn * 2];//判断每条边是否访问过。
int top;
int ans[maxn * 2];//保存路径
void dfs(int u)
{
    for (int i = head[u]; i != -1; i = edge[i].next)
    {
        if (!vis2[i])
        {
            vis2[i] = true;
            dfs(edge[i].to);
            ans[top++] = i;
        }
    }
}
int main()
{
    int T, n;
    scanf("%d", &T);
    while (T--)
    {
        init();
        scanf("%d", &n);
        memset(in, 0, sizeof(in));
        memset(out, 0, sizeof(out));
        memset(vis, false, sizeof(vis));
        int u, v;
        st = 10000;
        for (int i = 0; i <n; i++)
            cin >> str[i];
        sort(str, str + n);//从小到大排序
        for (int i = n - 1; i >= 0; i--)//因为链式前向星是逆序存图，所以反过来从大到小读入。
        {
            u = str[i][0] - ‘a‘ + 1;
            v = str[i][str[i].length() - 1] - ‘a‘ + 1;
            vis[u] = vis[v] = true;
            addedge(u, v, i);
            ++in[v];
            ++out[u];
            Union(u, v);
            st = min(st, min(u, v));//找出最小的那个点来
        }
        if (!check(st))
            puts("***");
        else
        {
            top = 0;
            memset(vis2, false, sizeof(vis2));
            dfs(st);
            for (int i = top - 1; i > 0; i--)
                cout << str[edge[ans[i]].id] << ".";
            cout << str[edge[ans[0]].id] << endl;
        }

    }
    return 0;
}